# How do you find the critical numbers for g(t) = t sqrt(4-t)  to determine the maximum and minimum?

Aug 3, 2018

PMAX(8/3;(16*sqrt(3))/9)

#### Explanation:

Given $g \left(t\right) = t \sqrt{4 - t}$
we get by the product and the chain rule

$g ' \left(t\right) = \sqrt{4 - t} + t \cdot \left(\frac{1}{2}\right) {\left(4 - t\right)}^{- \frac{1}{2}} \cdot \left(- 1\right)$
which simplifies to

$g ' \left(t\right) = \sqrt{4 - t} - \frac{t}{2 \sqrt{4 - t}}$
so we get

$g \cdot \left(t\right) = 0$ if

$\sqrt{4 - t} = \frac{t}{2 \sqrt{4 - t}}$ Multiplying with $2 \sqrt{4 - t}$

$2 \left(4 - t\right) = t$

so $t = \frac{8}{3}$

$g ' ' \left(t\right) = \frac{1}{2} {\left(4 - t\right)}^{- \frac{1}{2}} \cdot \left(- 1\right) - \frac{1}{2} \cdot {\left(4 - t\right)}^{- \frac{1}{2}} + \frac{t}{2} {\left(4 - t\right)}^{- \frac{3}{2}} \cdot \left(- \frac{1}{2}\right)$

$g ' ' \left(t\right) = \frac{t}{\sqrt{4 - t}} - \frac{t}{4 {\left(4 - t\right)}^{\frac{3}{2}}}$

$g ' ' \left(\frac{8}{3}\right) = - 3 \frac{\sqrt{3}}{4} < 0$
so we get PMAX(8/3;16*sqrt(3)/9)